Mathematics W4041x
Introduction to Modern Algebra
Answers to Midterm Exam #2
November 18, 2010
1. If G is a group, H and N subgroups with N C G, then HN is a subgroup, N C HN ,
H ∩ N C H, and H/(H ∩ N ) ∼
= HN/N .
2. If G 6∼
= 1, then there exists g ∈ G such that g 6= e. The subgroup hgi generated
by g is not 1, so it must be G. We know that every cyclic subgroup (that is, every
subgroup generated by a single element) is isomorphic either to Z or to Zn for some n
(Assignment 7 # 1). But Z has the nontrivial proper subgroup 2Z, while Zn for n = ab
composite has the nontrivial proper subgroup aZn . The only remaining possibility is
G∼
= Zp for p prime.
3. Suppose n ∈ N1 ∩ N2 and g ∈ G. Since N1 C G, gng −1 ∈ N1 . Likewise, since N2 C G,
gng −1 ∈ N2 . Hence gng −1 ∈ N1 ∩ N2 . So N1 ∩ N2 C G.
Alternative: Let π1 : G → G/N1 and π2 : G → G/N2 be the projection homomorphisms. Then N1 ∩ N2 = ker(π1 × π2 ) : G → G/N1 × G/N2 , and all kernels are
normal.
4. By the first isomorphism theorem, φ(G) ∼
= G/ ker φ, so #G = #φ(G)# ker φ and hence
#φ(G) | #G. And φ(G) is a subgroup of H, so by Lagrange’s theorem #φ(G) | #H
also. But (#G, #H) = 1, so #φ(G) = 1 and φ(G) = {e}.
5. Since x and y are in the same orbit, there exists g ∈ G such that y = g · x. Then
h ∈ Gy ⇔ h · y = y ⇔ h · (g · x) = g · x ⇔ g −1 · (h · (g · x)) = x ⇔ (g −1 hg) · x = x ⇔
g −1 hg ∈ Gx ⇔ h ∈ gGx g −1 .
6. Let φ1 (g) := φ(g, e) and φ2 (g) := φ(e, g). Then φ1 (gh) = φ(gh, e) = φ((g, e)(h, e)) =
φ(g, e)φ(h, e) = φ1 (g)φ2 (g), so φ1 is a homomorphism, and similarly for φ2 . And
φ(g1 , g2 ) = φ((g1 , e)(e, g2 )) = φ(g1 , e)φ(e, g2 ) = φ1 (g1 )φ2 (g2 ), but also φ(g1 , g2 ) =
φ((e, g2 )(g1 , e)) = φ(e, g2 )φ(g1 , e) = φ2 (g2 )φ1 (g1 ).
7. Let S be the set of all functions {1, 2, 3, 4, 5, 6, 7} → {B, W }. The colorings of the
necklace correspond to elements of this set. It is acted upon by the dihedral group
D14 , and we need to count the number of orbits. By Burnside’s lemma, this is given
by
X
1
#S g .
#D14 g∈D
14
The identity fixes all 27 functions. Since 7 is prime, any other rotation generates Z7 ,
which acts transitively on the set {1, 2, 3, 4, 5, 6, 7}. Hence the rotation fixes only 2
necklaces, the all-black and all-white ones. The other 7 elements are all reflections
in an axis through one bead, as shown. They fix the 24 necklaces having the same
colors on opposite pairs of beads. The grand total is therefore (27 + 6 · 2 + 7 · 24 )/14 =
(64 + 6 + 56)/7 = 18.